Directional Derivatives

[dcl]

Heya's
I need to find the direction in the xy plane in which one should travel, starting from point (1,1), to obtain the most rapid rate of decrease of
f(x,y,z) = (x + y - 2)^2 + (3x - y - 6)^2

now, \nabla f = (2(x+y-2), 2(3x-y-6))

so I'm thinking now I have to find the the unti vector 'u' which would be the direction in question. Unfortunately I do not know how to go on from here.
Somehow maximise \nabla f \cdot u (where 'u' is a unit vector, dunno how to do vectors properly in latex :frown: )

[Icarus]

\vec{u} (I looked it up myself just a few hours ago).

First of all, \nabla f = (2(x + y - 2) + 6(3x - y -6), 2(x + y - 2) - 2(3x - y - 6))
Second, you are only concerned with what is going on at (1,1), where \nabla f = (-24, 8).

Now, the question is to minimize (-24, 8)\cdot\vec{u}. Since \vec{v}\cdot\vec{w} = vw\cos \theta, where \theta is the angle between them, we see that the minimum occurs when \theta = \pi. I.e., when the vectors are pointing in opposite directions.

So \vec{u} must be the unital vector in the direction of (24, -8).

[dcl]

Silly me, I was straining all that time trying to figure out why my method is wrong and all the time I had been working with the wrong grad f. Thanks.